Nicodem

I’m not in the mood for answering so many questions so I will just tell how I see it (that is, I see it similar to the Ritter reaction*) so you can give answers yourself.

The first requirement for the reaction is that the alcohol forms a carbocation with the help of the sulphuric acid:1.) R3C-OH + H+ <==> R3C+ + H2O

This carbocation has then at least three ways to react further.

The first is the reverse reaction of the above, the nucleophylic addition of water formed above (alcohols can also add but we will ignore that as the reaction is analogous). This reaction is kept at its minimum with the water being removed by sulphuric acid:2.) R3C+ + H2O <==> R3C-OH + H+

The second is the beta proton elimination reaction:3.) R2C+-CH2-R <==> R2C=CH-R + H+

The third option is the addition of a nucleophyle. You want the product to bee an amine but ammonia or amines cannot be used due to the highly acidic environment needed for the reaction 1 to occur. Amines are fully protonated in sulphuric acid and as such are not nucleophyles. So you need a nitrogen source with a lone electron pair that will not get completely protonated at the reaction conditions, thus retaining some nucleophylicity. Urea (or R-CN in the Ritter reaction) is a really bad nucleophyle and as such does not posses much basicity (pKa=0.1) and is not completely protonated even by the strongest acids (a proton is also a very strong electrophyle). Thus, as a compromise, even though the nucleophylicity of urea is so low there is at least always some of it free (unprotonated) and this can add to the electrophylic carbocation (similarly like water in the reaction 2):4.) R3C+ + H2N-CO-NH2 ==> R3C-NH-CO-NH2 + H+

But given that both reactions 2 and 3 are fully reversible (alkenes also add a proton to the double bond forming back the tertiary carbocation) the urea product becomes the major product of the altogether reaction. This is because the urea group is a much less effective leaving group than the protonated alcohol group, making the reaction 4 the least reversible of all the other possible.

I hope this also makes it clear why this reaction works so well only with the tertiary alcohols and why concentrated sulphuric acid is used. The secondary carbocations are considerably harder to produce or put in other words their concentration in the reaction medium is possibly to low for a good reaction rate and satisfying yields.